生成与现有1D数组具有预先指定的相关性的NumPy 1D数组?

问题描述

我有一个未生成的1D NumPy数组.现在，我们将使用生成的一个.

I have a non-generated 1D NumPy array. For now, we will use a generated one.

import numpy as np

arr1 = np.random.uniform(0, 100, 1_000)

我需要一个与 0.3 相关的数组:

I need an array that will be correlated 0.3 with it:

arr2 = '?'
print(np.corrcoef(arr1, arr2))

Out[1]: 0.3

推荐答案

我已经通过用笨拙的方式解决了这个答案stats.SE到NumPy.想法是随机生成第二个数组 noise ，然后在 arr1 上计算 noise 的最小二乘线性回归的残差.残差与 arr1 的相关性必定为0，当然 arr1 与自身的相关性为1，因此 a * arr1 +的适当线性组合b *残基将具有任何所需的相关性.

I've adapted this answer by whuber on stats.SE to NumPy. The idea is to generate a second array noise randomly, and then compute the residuals of a least-squares linear regression of noise on arr1. The residuals necessarily have a correlation of 0 with arr1, and of course arr1 has a correlation of 1 with itself, so an appropriate linear combination of a*arr1 + b*residuals will have any desired correlation.

import numpy as np

def generate_with_corrcoef(arr1, p):
    n = len(arr1)

    # generate noise
    noise = np.random.uniform(0, 1, n)

    # least squares linear regression for noise = m*arr1 + c
    m, c = np.linalg.lstsq(np.vstack([arr1, np.ones(n)]).T, noise)[0]

    # residuals have 0 correlation with arr1
    residuals = noise - (m*arr1 + c)

    # the right linear combination a*arr1 + b*residuals
    a = p * np.std(residuals)
    b = (1 - p**2)**0.5 * np.std(arr1)

    arr2 = a*arr1 + b*residuals

    # return a scaled/shifted result to have the same mean/sd as arr1
    # this doesn't change the correlation coefficient
    return np.mean(arr1) + (arr2 - np.mean(arr2)) * np.std(arr1) / np.std(arr2)

最后一行缩放结果，以使平均值和标准偏差与 arr1 相同.但是， arr1 和 arr2 不会完全相同地分布.

The last line scales the result so that the mean and standard deviation are the same as arr1's. However, arr1 and arr2 will not be identically distributed.

用法:

>>> arr1 = np.random.uniform(0, 100, 1000)
>>> arr2 = generate_with_corrcoef(arr1, 0.3)
>>> np.corrcoef(arr1, arr2)
array([[1. , 0.3],
       [0.3, 1. ]])

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